A 1450 kg car moving south at 11.1 m/s collides with a 2460 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.01 m/s to the north. Find the velocity of the 2460 kg car before the collision.

Given two bodies A and B undergoing collision, using the principle of conservation of linear momentum that total momentum ([tex]P_{A1}[/tex] + [tex]P_{B1}[/tex]) of two bodies before collision is equal to the total momentum ([tex]P_{A2}[/tex] + [tex]P_{B2}[/tex]) of the bodies after collision. i.e

[tex]P_{A1}[/tex] + [tex]P_{B1}[/tex] = [tex]P_{A2}[/tex] + [tex]P_{B2}[/tex] ---------------(i)

Where;

[tex]P_{A1}[/tex] = Momentum of body A before collision

[tex]P_{B1}[/tex] = Momentum of body B before collision

[tex]P_{A2}[/tex] = Momentum of body A after collision

[tex]P_{B2}[/tex] = Momentum of body B after collision

Also;

[tex]P_{A1}[/tex] = [tex]m_{A}[/tex] x [tex]u_{A}[/tex] [ [tex]m_{A}[/tex] = mass of body A, [tex]u_{A}[/tex] = initial velocity of body A]

[tex]P_{B1}[/tex] = [tex]m_{B}[/tex] x [tex]u_{B}[/tex] [ [tex]m_{B}[/tex] = mass of body B, [tex]u_{B}[/tex] = initial velocity of body B]

[tex]P_{A2}[/tex] = [tex]m_{A}[/tex] x [tex]v_{A}[/tex] [ [tex]m_{A}[/tex] = mass of body A, [tex]u_{A}[/tex] = final velocity of body A]

[tex]P_{B2}[/tex] = [tex]m_{B}[/tex] x [tex]v_{B}[/tex] [ [tex]m_{B}[/tex] = mass of body B, [tex]v_{B}[/tex] = final velocity of body B]

Substitute these values into equation (i) as follows;

([tex]m_{A}[/tex] x [tex]u_{A}[/tex]) + ([tex]m_{B}[/tex] x [tex]u_{B}[/tex]) = ([tex]m_{A}[/tex] x [tex]v_{A}[/tex]) + ([tex]m_{B}[/tex] x [tex]v_{B}[/tex]) ------------------(ii)

But since the two move together after collision, then their final velocities are the same. i.e [tex]v_{A}[/tex] = [tex]v_{B}[/tex] = v

Therefore, equation (ii) becomes;

=> ([tex]m_{A}[/tex] x [tex]u_{A}[/tex]) + ([tex]m_{B}[/tex] x [tex]u_{B}[/tex]) = ([tex]m_{A}[/tex] x v) + ([tex]m_{B}[/tex] x v)

=> ([tex]m_{A}[/tex] x [tex]u_{A}[/tex]) + ([tex]m_{B}[/tex] x [tex]u_{B}[/tex]) = v ([tex]m_{A}[/tex] + [tex]m_{B}[/tex] ) ----------------------------(iii)

Let south direction be negative and north direction be positive;

Therefore, from question;

[tex]m_{A}[/tex] = 1450kg

[tex]u_{A}[/tex] = - 11.1m/s [south is negative]

[tex]m_{B}[/tex] = 2460kg

[tex]u_{B}[/tex] = ?

v = +5.01m/s [north is positive]

Substitute these values into equation (iii) as follows;

=> (1450 x -11.1) + (2460 x [tex]u_{B}[/tex]) = 5.01(1450 + 2460)

=> (-16095) + (2460[tex]u_{B}[/tex]) = 19589.1

=> -16095 + 2460[tex]u_{B}[/tex] = 19589.1

=> 2460[tex]u_{B}[/tex] = 19589.1 + 16095

=> 2460[tex]u_{B}[/tex] = 35684.1

=> [tex]u_{B}[/tex] = 35684.1 / 2460

=> [tex]u_{B}[/tex] = 14.51 m/s

Therefore, the velocity of body B (second car with 2460kg) before collision is 14.51m/s and since it is positive it shows that it is in the north direction which confirms the situation in the question.